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A generalized lasso formulation of weighted and regularized (nearly-)isotonic regression on graphs

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Abstract

In this paper we study a regularized version of nearly-isotonic regression with respect to a general partial order. In this estimator the sum of squares is weighted, with some of the weights allowed to be zero, and an additional quadratic penalty is imposed. The only assumption we make is that the quadratic part of the objective function is strictly convex, and we show that this assumption is mild. First, we show that for all sufficiently large values of the penalty parameter the estimator coincides with the regularized isotonic regression, in which the penalty is replaced by the isotonic constraint. Further, we prove that the resulting problem can be written as a generalized lasso problem with a design matrix of full column rank and with the penalty matrix unchanged. Next, we derive the dual problem, which is a quadratic program with box constraints, and write it in a sparse form that can be solved by standard quadratic programming solvers for sparse problems.

Introduction

In the area of constrained inference the basic and simplest problem is isotonic regression, i.e. the ℓ2\ell_2-projection of a given vector of data points y∈Rn\mathbf{y} \in\mathbb{R}^n onto the cone of non-decreasing vectors in Rn\mathbb{R}^n,

β^I=arg min⁡β∈Rn ∥y−β∥22,subject to β1≤β2≤⋯≤βn.(1)\widehat{\beta}^{I} = \underset{\beta\in\mathbb{R}^n}{\operatorname{arg\,min}}\ \lVert\mathbf{y} - \beta\rVert_2^2,\qquad\text{subject to }\beta_1 \le\beta_2 \le\cdots\le\beta_n. \tag*{(1)}

A general introduction can be found in [14]. The nearly-isotonic regression, introduced in [20] and studied in detail in [11], is the less restrictive version

β^NI(y,λNI)=arg min⁡β∈Rn 12∥y−β∥22+λNI∑i=1n−1(βi−βi+1)+,(2)\widehat{\beta}^{NI}(\mathbf{y},\lambda_{NI}) = \underset{\beta\in\mathbb{R}^n}{\operatorname{arg\,min}}\ \frac{1}{2}\lVert\mathbf{y} - \beta\rVert_2^2 + \lambda_{NI}\sum_{i=1}^{n-1}(\beta_i-\beta_{i+1})_+, \tag*{(2)}

where (t)+=max⁡{t,0}(t)_+ = \max\{t,0\}, in which the constraint is replaced by a penalty on its violations. Its combination with the fused lasso and the extension to a general partial order were studied in [12], where in particular the dual of the problem was shown to be a quadratic program with box constraints.

The subject of penalized estimation starts with the ℓ1\ell_1-penalisation, i.e. the lasso [18], and the ℓ2\ell_2-penalisation, i.e. ridge regression [9]. The ℓ1\ell_1-penalisation of the successive differences of the coefficients gives the fusion approximator [13], and its combination with the lasso is the fused lasso [19]. These estimators were studied and solved for a general graph in [4, 8, 21]. In [21] they appear as special cases of the generalized lasso, in which the ℓ1\ell_1-penalty is imposed on DβD\beta for a general penalty matrix DD, and trend filtering on graphs [22] is another example. The lasso with linear constraints on the coefficients, called the constrained lasso, and its relation to the generalized lasso were studied in [5]. There the constraint matrices are assumed to have full row rank, and for isotonic regression this holds only if the graph of the partial order is a tree, which is not the case, for example, for the bimonotone order on a grid. Next, the statistical properties of isotonic regression in general dimensions were studied in [3, 7], and algorithms for it can be found, for example, in [2, 16]. Several regularized versions of isotonic regression have been proposed. Generalizations of nearly-isotonic regression in the one-dimensional case were studied in [23], where a dynamic programming algorithm is proposed, and in [6], where the squares of the positive parts of the differences and a lasso penalty are added to the penalty in (2).

In this paper we study the nearly-isotonic regression with respect to a general partial order in the following modified form: the sum of squares is weighted, with some of the weights possibly equal to zero, and a quadratic penalty is added. This modification is motivated by [1], where the authors study least squares estimation under bimonotonicity constraints for the data observed on an incomplete grid. There the weights are the numbers of observations at the grid points, and they are equal to zero at the empty cells. In this case the problem is not strictly convex, and the authors add a small quadratic penalty, which they call light regularization. The quadratic penalty is also the one of the smoothed monotonic regression of [17] in the one-dimensional case. The main result of the paper is that the modified problem can be written as a generalized lasso problem in the sense of [21], with the same penalty matrix and with a design matrix of full column rank, and can therefore be solved by the algorithms developed for the generalized lasso. We also derive the dual problem.

Notation

We start with basic definitions of partial order and isotonic regression. Let I={i1,…,in}I = \{i_1,\ldots,i_n\} be some index set. A binary relation ⪯\preceq on II is called partial order if

  • it is reflexive, i.e. j⪯jj \preceq j for all j∈Ij \in I,

  • it is transitive, i.e. j1,j2,j3∈Ij_1,j_2,j_3 \in I, j1⪯j2j_1 \preceq j_2 and j2⪯j3j_2 \preceq j_3 imply j1⪯j3j_1 \preceq j_3,

  • it is antisymmetric, i.e. j1,j2∈Ij_1,j_2 \in I, j1⪯j2j_1 \preceq j_2 and j2⪯j1j_2 \preceq j_1 imply j1=j2j_1 = j_2.

Further, a vector β∈Rn\beta\in\mathbb{R}^n indexed by II is called isotonic with respect to the partial order ⪯\preceq on II if j1⪯j2j_1 \preceq j_2 implies βj1≤βj2\beta_{j_1} \le\beta_{j_2}. We denote the set of all isotonic vectors in Rn\mathbb{R}^n with respect to ⪯\preceq by BI\mathcal{B}^I. This set is a closed convex cone in Rn\mathbb{R}^n, and it is also called the isotonic cone. Next, the isotonic regression of a vector y∈Rny \in\mathbb{R}^n indexed by the partially ordered index set II is given by

β^I=arg⁡min⁡β∈BI∑j∈I(βj−yj)2(3)\hat{\beta}^{I} = \arg\min_{\beta\in\mathcal{B}^{I}} \sum_{j \in I}(\beta_j-y_j)^2 \tag*{(3)}

which for the natural order on I={1,…,n}I = \{1,\ldots,n\} is (1).

For any partial order relation ⪯\preceq on II there exists a directed graph G=(V,E)G=(V,E), with V=IV=I and EE the minimal set of edges (j1,j2)(j_1,j_2) with j1⪯j2j_1 \preceq j_2, such that β\beta is isotonic with respect to ⪯\preceq if and only if βl1≤βl2\beta_{l_1} \le\beta_{l_2} whenever EE contains a chain of edges from l1l_1 to l2l_2. We label the vertices by {1,…,n}\{1,\ldots,n\} and the edges by {1,…,m}\{1,\ldots,m\}, and let D∈Rm×nD \in\mathbb{R}^{m \times n} be the oriented incidence matrix of GG,

De,j={1,if vertex j is the source of edge e,−1,if vertex j is the target of edge e,0,otherwise,(4)D_{e,j} = \begin{cases} 1, & \text{if vertex } j \text{ is the source of edge } e,\\ -1, & \text{if vertex } j \text{ is the target of edge } e,\\ 0, & \text{otherwise,} \end{cases} \tag*{(4)}

so that DβD\beta is the vector of the differences of β\beta along the edges of GG. Therefore, the isotonic cone is

Bis={β∈Rn:Dβ≤0},\mathcal{B}^{\mathrm{is}} = \{\beta\in\mathbb{R}^{n} : D\beta\le0\},

where the inequality is meant coordinatewise. Further, for z∈Rmz \in\mathbb{R}^{m} we write

∥z∥+=∑e=1m(ze)+.\lVert z\rVert_{+} = \sum_{e=1}^{m}(z_e)_{+}.

For the natural order on I={1,…,n}I=\{1,\ldots,n\} the edges join consecutive indices, and ∥Dβ∥+\lVert D\beta\rVert_{+} is the penalty in (2). For the bimonotone order on a grid, introduced in [1], the edges join horizontally and vertically adjacent grid points.

General statement of the problem

Let y∈Rny \in\mathbb{R}^{n} be a signal indexed by II, let WW be a diagonal matrix with nonnegative weights w1,…,wnw_1,\ldots,w_n on the diagonal, not all equal to zero, let SS be a given matrix with nn columns and ℓ\ell rows, and let λNI≥0\lambda_{NI} \ge0. We consider the estimator

β^=arg min⁡β∈Rn Φ(β),Φ(β)=(y−β)TW(y−β)+∥Sβ∥22+λNI∥Dβ∥+,(5)\widehat{\beta} = \underset{\beta\in\mathbb{R}^{n}}{\operatorname{arg\,min}}\ \Phi(\beta), \qquad\Phi(\beta) = (y-\beta)^{\mathsf{T}}W(y-\beta) + \lVert S\beta\rVert_{2}^{2} + \lambda_{NI}\lVert D\beta\rVert_{+}, \tag*{(5)}

under the assumption that

β↦(y−β)TW(y−β)+∥Sβ∥22\beta\mapsto(y-\beta)^{\mathsf{T}}W(y-\beta) + \lVert S\beta\rVert_{2}^{2}

is strictly convex.

λNI∑e=(i,j)∈Eωe(βi−βj)+=λNI∥Dωβ∥+,Dω=diag⁡(ω1,…,ωm)D.\lambda_{NI}\sum_{e=(i,j)\in E}\omega_e(\beta_i-\beta_j)_{+} = \lambda_{NI}\lVert D_{\omega}\beta\rVert_{+}, \qquad D_{\omega} = \operatorname{diag}(\omega_1,\ldots,\omega_m)D.

Theorem 2 and Propositions 3 and 6 below hold with DD replaced by DωD_{\omega}, since their proofs do not use the structure of DD.

The structure of (5) is that of the “incomplete layout with light regularization” in Setting 2b of [1]. There, the observations of a bimonotone regression function are reduced to a rectangular grid: wuw_u is the number of observations at the grid point uu and ZuZ_u their average, and the least squares criterion becomes

∑uwu(Zu−θu)2.\sum_u w_u(Z_u-\theta_u)^2.

When the layout is incomplete, i.e. some grid points have no observations and wu=0w_u=0, this function is not strictly convex, and the least squares fit is not uniquely determined at the empty cells. For this reason the authors consider the minimisation of

∑uwu(Zu−θu)2+λSP(θ)\sum_u w_u(Z_u-\theta_u)^2+\lambda_S P(\theta)

over the bimonotone cone, where PP is a convex quadratic function chosen so that the sum is strictly convex, and λS>0\lambda_S>0 is small, its role being only to make the Hessian well conditioned.

The authors discuss two choices of PP, the Tychonov regularisation with a reference value θ0\theta_0 and the sum of squared differences along the edges of the grid,

P(θ)=∑u(θu−θ0)2andP(θ)=∑(u,v)(θv−θu)2.P(\theta) = \sum_u(\theta_u-\theta_0)^2 \qquad\text{and} \qquad P(\theta) = \sum_{(u,v)}(\theta_v-\theta_u)^2.

They prefer the latter, because it interpolates more smoothly into the empty cells. In the notation of (5) the latter penalty is ∥Sβ∥22\lVert S\beta\rVert_2^2 with

S=λSD,(6)S = \sqrt{\lambda_S}D, \tag*{(6)}

and their requirement on PP is the assumption (6). In Lemma 4 below we verify (6) for this choice. Problem (5) is thus the nearly-isotonic relaxation of the problem of [1], in which the isotonic constraint is replaced by the penalty, and the bimonotone order is replaced by an arbitrary partial order. On the chain graph, with unit weights and with the constraint kept, the same quadratic part gives the smoothed monotonic regression of [17] with equal penalty parameters.

Organisation of the paper

The rest of the paper is organised as follows. In Section 2 we relate the estimator to the constrained estimator in which the penalty λNI∥Dβ∥+\lambda_{NI}\lVert D\beta\rVert_+ is replaced by the constraint Dβ≤0D\beta\le0. In Section 3 we show that (5) is a generalized lasso problem. Section 4 contains the dual problem. In Section 5 we verify assumption (6) for the penalty (7). Section 6 is devoted to the computational aspects. The paper closes with a discussion in Section 7.

Relation to regularized isotonic regression

In this section we compare the estimator (5) with the estimator obtained when the penalty λNI∥Dβ∥+\lambda_{NI}\lVert D\beta\rVert_+ in (5) is removed and the isotonic constraint Dβ≤0D\beta\le0 is imposed instead,

β^CIS=arg⁡min⁡β∈Rn  (y−β)TW(y−β)+∥Sβ∥22subject toDβ≤0.(7)\widehat{\beta}^{CIS} = \underset{\beta\in\mathbb{R}^n}{\arg\min}\;(\boldsymbol{y}-\beta)^{\mathsf{T}}W(\boldsymbol{y}-\beta)+\lVert S\beta\rVert_2^2 \qquad\text{subject to}\qquad D\beta\le0. \tag*{(7)}

For the grid and S=λSDS=\sqrt{\lambda_S}D this is the estimator of [1], and for the chain it is the smoothed monotonic regression of [17]. For nearly-isotonic regression, i.e. for W=12InW=\frac{1}{2}I_n and S=0S=0, the property that the estimator coincides with isotonic regression for all sufficiently large λNI\lambda_{NI} is mentioned in [11] and [20]. The same holds for the estimator (5). Under (6) the objective functions in (5) and (8) are strictly convex and coercive, so both problems have unique solutions. Further, the objective function in (5) is the exact penalty function of the problem (8) with the penalty parameter λNI\lambda_{NI}, and the constraints in (8) are linear. Therefore, by the classical exact penalty result, see [10] and the references therein, there exists λ0≥0\lambda_0\ge0 such that β^=β^CIS\widehat{\beta}=\widehat{\beta}^{CIS} for all λNI≥λ0\lambda_{NI}\ge\lambda_0, in particular, for nearly-isotonic regression with respect to a general partial order.

Thus, the estimator (5) is the regularized isotonic regression (8) for all sufficiently large values of λNI\lambda_{NI}, and the results of the following sections apply to the estimator (8) as well.

Solution via the generalized lasso

First, we rewrite the objective function in (5).

Lemma 1. For a fixed data vector y∈Rn\mathbf{y} \in\mathbb{R}^{n} indexed by II the objective function in (5) can be written as

Φ(β)=βTQβ−2bTβ+λNI2∥Dβ∥1+yTWy,(8)\Phi(\boldsymbol{\beta}) = \boldsymbol{\beta}^{\mathsf{T}}Q\boldsymbol{\beta} - 2\mathbf{b}^{\mathsf{T}}\boldsymbol{\beta} + \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1} + \mathbf{y}^{\mathsf{T}}W\mathbf{y}, \tag*{(8)}

where

Q=W+STS∈Rn×n,b=Wy−λNI4DT1∈Rn,(9)Q = W + S^{\mathsf{T}}S \in\mathbb{R}^{n \times n}, \qquad\mathbf{b} = W\mathbf{y} - \frac{\lambda_{NI}}{4}D^{\mathsf{T}}\mathbf{1} \in\mathbb{R}^{n}, \tag*{(9)}

and 1∈Rm\mathbf{1} \in\mathbb{R}^{m} is the vector whose all elements are equal to one. Further, the assumption (6) holds if and only if QQ is positive definite.

Proof. The function in (6) is a quadratic function with constant Hessian 2Q2Q, and a quadratic function is strictly convex if and only if its Hessian is positive definite. This gives the last statement of the lemma.

For (9), since WW is symmetric, (y−β)TW(y−β)=βTWβ−2(Wy)Tβ+yTWy(\mathbf{y} - \boldsymbol{\beta})^{\mathsf{T}}W(\mathbf{y} - \boldsymbol{\beta}) = \boldsymbol{\beta}^{\mathsf{T}}W\boldsymbol{\beta} - 2(W\mathbf{y})^{\mathsf{T}}\boldsymbol{\beta} + \mathbf{y}^{\mathsf{T}}W\mathbf{y}, and ∥Sβ∥22=βTSTSβ\lVert S\boldsymbol{\beta}\rVert_{2}^{2} = \boldsymbol{\beta}^{\mathsf{T}}S^{\mathsf{T}}S\boldsymbol{\beta}. Together these give βTQβ−2(Wy)Tβ+yTWy\boldsymbol{\beta}^{\mathsf{T}}Q\boldsymbol{\beta} - 2(W\mathbf{y})^{\mathsf{T}}\boldsymbol{\beta} + \mathbf{y}^{\mathsf{T}}W\mathbf{y}. For t∈Rt \in\mathbb{R} one has (t)+=12∣t∣+12t(t)_{+} = \frac{1}{2}|t| + \frac{1}{2}t. Therefore,

λNI∥Dβ∥+=λNI2∥Dβ∥1+λNI21TDβ=λNI2∥Dβ∥1+2(λNI4DT1)Tβ,\lambda_{NI}\lVert D\boldsymbol{\beta}\rVert_{+} = \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1} + \frac{\lambda_{NI}}{2}\mathbf{1}^{\mathsf{T}}D\boldsymbol{\beta} = \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1} + 2\left(\frac{\lambda_{NI}}{4}D^{\mathsf{T}}\mathbf{1}\right)^{\mathsf{T}}\boldsymbol{\beta},

which finishes the proof. □\square

Next, we provide the solution to the problem (5).

Theorem 2. Assume (6) holds, let QQ and b\mathbf{b} be as in (10), let W1/2=diag⁡(w1,…,wn)W^{1/2} = \operatorname{diag}(\sqrt{w_{1}}, \ldots, \sqrt{w_{n}}), and let

X=2(W1/2S)∈R(n+ℓ)×n,y~=XQ−1b∈Rn+ℓ.(10)X = \sqrt{2}\begin{pmatrix} W^{1/2} \\ S \end{pmatrix} \in\mathbb{R}^{(n+\ell)\times n}, \qquad\widetilde{\mathbf{y}} = XQ^{-1}\mathbf{b} \in\mathbb{R}^{n+\ell}. \tag*{(10)}

Then the solution to the problem (5) is given by

β^=arg min⁡β∈Rn{12∥y~−Xβ∥22+λNI2∥Dβ∥1}.(11)\widehat{\boldsymbol{\beta}} = \underset{\boldsymbol{\beta}\in\mathbb{R}^{n}}{\operatorname{arg\,min}}\left\{\frac{1}{2}\lVert\widetilde{\mathbf{y}} - X\boldsymbol{\beta}\rVert_{2}^{2} + \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1}\right\}. \tag*{(11)}

That is, the problem (5) is the generalized lasso problem with the data vector y~\widetilde{\mathbf{y}}, the design matrix XX, the penalty matrix DD and the penalty parameter λNI2\frac{\lambda_{NI}}{2}. Moreover, XX has full column rank and the solution (12) is unique.

Proof. First, note that

XTX=2W1/2W1/2+2STS=2QandXTy~=XTXQ−1b=2b.X^{\mathsf{T}}X = 2W^{1/2}W^{1/2} + 2S^{\mathsf{T}}S = 2Q \qquad\text{and} \qquad X^{\mathsf{T}}\widetilde{\mathbf{y}} = X^{\mathsf{T}}XQ^{-1}\mathbf{b} = 2\mathbf{b}.

Next, from Lemma 1 it follows that

Φ(β)=βTQβ−2bTβ+λNI2∥Dβ∥1+yTWy=12βTXTXβ−y~TXβ+λNI2∥Dβ∥1+yTWy=12∥y~−Xβ∥22+λNI2∥Dβ∥1+yTWy−12∥y~∥22.\begin{aligned} \Phi(\boldsymbol{\beta}) &= \boldsymbol{\beta}^{\mathsf{T}}Q\boldsymbol{\beta} - 2\mathbf{b}^{\mathsf{T}}\boldsymbol{\beta} + \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1} + \mathbf{y}^{\mathsf{T}}W\mathbf{y} \\ &= \frac{1}{2}\boldsymbol{\beta}^{\mathsf{T}}X^{\mathsf{T}}X\boldsymbol{\beta} - \widetilde{\mathbf{y}}^{\mathsf{T}}X\boldsymbol{\beta} + \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1} + \mathbf{y}^{\mathsf{T}}W\mathbf{y} \\ &= \frac{1}{2}\lVert\widetilde{\mathbf{y}} - X\boldsymbol{\beta}\rVert_{2}^{2} + \frac{\lambda_{NI}}{2}\lVert D\boldsymbol{\beta}\rVert_{1} + \mathbf{y}^{\mathsf{T}}W\mathbf{y} - \frac{1}{2}\lVert\widetilde{\mathbf{y}}\rVert_{2}^{2}. \end{aligned}

Therefore, the problems (5) and (12) have the same solutions, since the last two terms do not depend on β\boldsymbol{\beta}. Lastly, by Lemma 1 the matrix QQ is positive definite, so is XTX=2QX^{\mathsf{T}}X = 2Q, which means that XX has full column rank and the objective function in (12) is strictly convex and coercive. Therefore, the solution to (12) exists and is unique. □\square

The pair XX, y~\widetilde{y} in (10) is not the only possible choice. As the proof shows, the statement of Theorem 2 holds for any X∈Rp×nX \in\mathbb{R}^{p \times n} and y~∈Rp\widetilde{y} \in\mathbb{R}^{p} such that

XTX=2QandXTy~=2b.(12)X^{\mathsf{T}}X = 2Q \qquad\text{and} \qquad X^{\mathsf{T}}\widetilde{y} = 2b. \tag*{(12)}

For example, (13) holds for the square matrix XX and the vector y~\widetilde{y} given by

X=2Q1/2,y~=2Q−1/2b.X = \sqrt{2}Q^{1/2}, \qquad\widetilde{y} = \sqrt{2}Q^{-1/2}b.

Finally, for nearly-isotonic regression, i.e. for W=12InW = \frac{1}{2}I_n and S=0S = 0, (13) holds for

X=In,y~=y−λNI2DT1,X = I_n, \qquad\widetilde{y} = y - \frac{\lambda_{NI}}{2}D^{\mathsf{T}}\mathbf{1},

and (12) becomes the relation

β^NI(y,λNI)=β^F(y−λNI2DT1,12λNI)\widehat{\beta}^{NI}(y,\lambda_{NI}) = \widehat{\beta}^{F}\left(y-\frac{\lambda_{NI}}{2}D^{\mathsf{T}}\mathbf{1},\frac{1}{2}\lambda_{NI}\right)

between nearly-isotonic regression and the fusion approximator, proved in [12].

The dual problem

For a fixed λNI\lambda_{NI} we can also solve the problem (5) numerically through its dual, which is a quadratic program with box constraints.

Proposition 3. Assume (6) holds. The solution to (5) is given by

β^=y~d−D~Tu^,(13)\widehat{\beta} = \widetilde{y}_d - \widetilde{D}^{\mathsf{T}}\widehat{u}, \tag*{(13)}

where

y~d=Q−1Wy∈Rn,D~=12DQ−1∈Rm×n,(14)\widetilde{y}_d = Q^{-1}Wy \in\mathbb{R}^{n}, \qquad\widetilde{D} = \frac{1}{2}DQ^{-1} \in\mathbb{R}^{m \times n}, \tag*{(14)}

and

u^∈arg⁡min⁡u∈Rm∥y~d−D~Tu∥Q2s.t.0≤u≤λNI1,(15)\widehat{u} \in\mathop{\arg\min}_{u \in\mathbb{R}^{m}} \left\lVert\widetilde{y}_d - \widetilde{D}^{\mathsf{T}}u \right\rVert_{Q}^{2} \qquad\text{s.t.} \qquad0 \leq u \leq\lambda_{NI}\mathbf{1}, \tag*{(15)}

with ∥v∥Q2=vTQv\lVert v\rVert_{Q}^{2} = v^{\mathsf{T}}Qv for v∈Rnv \in\mathbb{R}^{n}.

Proof. First, following [21], we write the problem (5) in the following way

minimize⁡β,z(y−β)TW(y−β)+∥Sβ∥22+λNI∥z∥+subject to⁡Dβ=z∈Rm.\begin{aligned} \underset{\beta,z}{\operatorname{minimize}}\quad&(y-\beta)^{\mathsf{T}}W(y-\beta)+\lVert S\beta\rVert_{2}^{2}+\lambda_{NI}\lVert z\rVert_{+}\\ \operatorname{subject\ to}\quad&D\beta=z\in\mathbb{R}^{m}. \end{aligned}

Further, the Lagrangian is given by

L(β,z,u)=(y−β)TW(y−β)+∥Sβ∥22+λNI∥z∥++uT(Dβ−z),L(\beta,z,u)=(y-\beta)^{\mathsf{T}}W(y-\beta)+\lVert S\beta\rVert_{2}^{2}+\lambda_{NI}\lVert z\rVert_{+}+u^{\mathsf{T}}(D\beta-z),

where u∈Rmu \in\mathbb{R}^{m} is a dual variable. Note that

inf⁡z(λNI∥z∥+−uTz)={0,if 0≤u≤λNI1,−∞,otherwise,\inf_{z}\left(\lambda_{NI}\lVert z\rVert_{+}-u^{\mathsf{T}}z\right)= \begin{cases} 0, & \text{if } 0\leq u\leq\lambda_{NI}\mathbf{1},\\ -\infty, & \text{otherwise}, \end{cases}

and, since QQ is positive definite,

min⁡β((y−β)TW(y−β)+∥Sβ∥22+uTDβ)=−14(2Wy−DTu)TQ−1(2Wy−DTu)+yTWy=−∥y~d−D~Tu∥Q2+yTWy.\begin{aligned} \min_{\beta}\left((y-\beta)^{\mathsf{T}}W(y-\beta)+\lVert S\beta\rVert_{2}^{2}+u^{\mathsf{T}}D\beta\right) &=-\frac{1}{4}(2Wy-D^{\mathsf{T}}u)^{\mathsf{T}}Q^{-1}(2Wy-D^{\mathsf{T}}u)+y^{\mathsf{T}}Wy\\ &=-\left\lVert\widetilde{y}_d-\widetilde{D}^{\mathsf{T}}u\right\rVert_{Q}^{2}+y^{\mathsf{T}}Wy. \end{aligned}

Next, the dual function is given by

g(u)=inf⁡β,zL(β,z,u)={−∥y~d−D~Tu∥Q2+yTWy,if 0≤u≤λNI1,−∞,otherwise,g(u) = \inf_{\beta,z} L(\beta,z,u) = \begin{cases} -\lVert\widetilde{y}_d-\widetilde{D}^{\mathsf{T}}u\rVert_Q^2+y^{\mathsf{T}}Wy, & \text{if } 0\le u\le\lambda_{NI}1,\\ -\infty, & \text{otherwise}, \end{cases}

and, therefore, the dual problem is

u^∈arg max⁡u g(u)subject to0≤u≤λNI1,\widehat{u}\in\underset{u}{\operatorname{arg\,max}}\ g(u)\quad\text{subject to}\quad0\le u\le\lambda_{NI}1,

which is equivalent to (15). Lastly, since the objective function is finite and convex on Rn×Rm\mathbb{R}^{n}\times\mathbb{R}^{m} and the constraint Dβ=zD\beta=z is affine and feasible, strong duality holds, and taking first derivative of the Lagrangian L(β,z,u^)L(\beta,z,\widehat{u}) with respect to β\beta we get the following relation between β^\widehat{\beta} and u^\widehat{u}

β^=y~d−D~Tu^.\widehat{\beta}=\widetilde{y}_d-\widetilde{D}^{\mathsf{T}}\widehat{u}.

If DD does not have full row rank, the minimiser in (15) need not be unique. However, the objective function in (15) is a strictly convex function of DTuD^{\mathsf{T}}u, therefore, DTu^D^{\mathsf{T}}\widehat{u} is unique, and (13) does not depend on the choice of u^\widehat{u}. Note also that y~d\widetilde{y}_d is the solution to (5) for λNI=0\lambda_{NI}=0, so (13) represents the estimator as the quadratic smoother minus a correction determined by the dual problem. For nearly-isotonic regression we have y~d=y\widetilde{y}_d=y and D~=D\widetilde{D}=D, and Proposition 3 becomes Theorem 1 of [12] without the fusion penalty.

The strict convexity assumption

In this section we verify the assumption (6) for the penalty (6), i.e. for S=λSDS=\sqrt{\lambda_S}D.

Lemma 4. Let the undirected graph underlying GG be connected, and let S=λSDS=\sqrt{\lambda_S}D with λS>0\lambda_S>0. Then the assumption (6) holds if and only if wi>0w_i>0 for at least one i∈{1,…,n}i\in\{1,\ldots,n\}.

Proof. With S=λSDS=\sqrt{\lambda_S}D we have Q=W+λSDTDQ=W+\lambda_SD^{\mathsf{T}}D, and by Lemma 1 the assumption (6) holds if and only if QQ is positive definite. Since WW is diagonal and vTDTDv=∥Dv∥22v^{\mathsf{T}}D^{\mathsf{T}}Dv=\lVert Dv\rVert_2^2, for every v∈Rnv\in\mathbb{R}^{n} we have

vTQv=∑i=1nwivi2+λS∥Dv∥22,(16)v^{\mathsf{T}}Qv=\sum_{i=1}^{n}w_iv_i^2+\lambda_S\lVert Dv\rVert_2^2, \tag*{(16)}

and both summands are nonnegative, because wi≥0w_i\ge0 and λS>0\lambda_S>0.

First, let wi>0w_i>0 for some ii. By (4), Dv=0Dv=0 if and only if vi=vjv_i=v_j for every edge (i,j)∈E(i,j)\in E, and since the undirected graph underlying GG is connected, this holds if and only if v=t1nv=t1_n for some t∈Rt\in\mathbb{R}, where 1n∈Rn1_n\in\mathbb{R}^{n} is the vector of ones. Now let v≠0v\ne0. If Dv≠0Dv\ne0, the second summand in (16) is strictly positive. If Dv=0Dv=0, then v=t1nv=t1_n with t≠0t\ne0, and the first summand in (16) equals t2∑i=1nwi>0t^2\sum_{i=1}^{n}w_i>0. Hence, vTQv>0v^{\mathsf{T}}Qv>0 for all v≠0v\ne0, i.e. QQ is positive definite.

Second, let wi=0w_i=0 for all ii. Then for v=1nv=1_n we have D1n=0D1_n=0, because every row of DD contains exactly one 11 and one −1-1, and therefore vTQv=0v^{\mathsf{T}}Qv=0 by (16). Hence QQ is not positive definite.

Therefore, for a connected graph the standing assumption that not all weights are equal to zero is necessary and sufficient. In particular, for the bimonotone grid, which is connected, the assumption (6) holds for every incomplete layout with at least one observation and for every λS>0\lambda_S>0. Note that without the quadratic penalty, i.e. for S=0S=0, the Hessian of the quadratic part of (5) is 2W2W, which is singular as soon as wi=0w_i=0 for some ii.

If the graph is not connected, then for S=λSDS=\sqrt{\lambda_S}D all terms in (5) are sums over the connected components, because no edge joins two different components. Therefore, the problem (5) splits into independent problems, one for each component, and Lemma 4 can be applied to each of them.

Corollary 5. Let S=λSDS=\sqrt{\lambda_S}D with λS>0\lambda_S>0. Then the assumption (6) holds if and only if every connected component of the undirected graph underlying GG contains a vertex ii with wi>0w_i>0.

Proof. Let C1,…,CKC_1,\ldots,C_K be the connected components of the undirected graph underlying GG. Since no edge joins two different components, after a permutation of the vertices the matrix Q=W+λSDTDQ=W+\lambda_SD^{\mathsf T}D is block diagonal with the blocks Qk=Wk+λSDkTDkQ_k=W_k+\lambda_SD_k^{\mathsf T}D_k, k=1,…,Kk=1,\ldots,K, where WkW_k is the diagonal matrix of the weights wiw_i, i∈Cki\in C_k, and DkD_k is the incidence matrix of the subgraph of GG with the vertex set CkC_k. Therefore, QQ is positive definite if and only if every QkQ_k is positive definite, and the statement follows from Lemma 4 applied to each component. □\square

Finally, the Tychonov penalty of [1] shrinks the estimator towards the reference value θ0\theta_0, and for θ0≠0\theta_0\ne0 it is not of the form ∥Sβ∥22\lVert S\beta\rVert_2^2. Nevertheless, the corresponding problem

min⁡β∈Rn(y−β)TW(y−β)+λS∑i=1n(βi−θ0)2+λNI∥Dβ∥+\min_{\beta\in\mathbb{R}^n}(\boldsymbol{y}-\boldsymbol{\beta})^{\mathsf T}W(\boldsymbol{y}-\boldsymbol{\beta})+\lambda_S\sum_{i=1}^{n}(\beta_i-\theta_0)^2+\lambda_{NI}\lVert D\boldsymbol{\beta}\rVert_+

reduces to (5). Indeed, let β′=β−θ01n\boldsymbol{\beta}'=\boldsymbol{\beta}-\theta_0\boldsymbol{1}_n and y′=y−θ01n\boldsymbol{y}'=\boldsymbol{y}-\theta_0\boldsymbol{1}_n. Then

y−β=y′−β′,∑i=1n(βi−θ0)2=∥β′∥22,Dβ=Dβ′,\boldsymbol{y}-\boldsymbol{\beta}=\boldsymbol{y}'-\boldsymbol{\beta}',\qquad\sum_{i=1}^{n}(\beta_i-\theta_0)^2=\lVert\boldsymbol{\beta}'\rVert_2^2,\qquad D\boldsymbol{\beta}=D\boldsymbol{\beta}',

where the last equality holds because D1n=0D\boldsymbol{1}_n=0. Therefore, the solution is

β^=β^′+θ01n,\widehat{\boldsymbol{\beta}}=\widehat{\boldsymbol{\beta}}'+\theta_0\boldsymbol{1}_n,

where β^′\widehat{\boldsymbol{\beta}}' is the solution to (5) with the data vector y′\boldsymbol{y}' and S=λSInS=\sqrt{\lambda_S}I_n. For this choice of SS the matrix QQ is positive definite for any weights, so the assumption (6) always holds.

Computational aspects

In this section we discuss the computation of the estimator (5). We do not provide a simulation study, because the algorithms below are well established, and their performance has been studied extensively, see [12] for the nearly-isotonic case, and also [15], [21].

First, by Theorem 2 the estimator is the solution to the generalized lasso problem (11). Therefore, we can use the algorithm of [21] for the problem (5), for an arbitrary partial order, and the condition of full column rank of the design matrix, required there, is satisfied automatically. Note, that the data vector y~\widetilde{\boldsymbol{y}} depends on λNI\lambda_{NI}, therefore, the path of [21] in the lasso parameter with fixed y~\widetilde{\boldsymbol{y}} is not the path of (5) in λNI\lambda_{NI}. Further, among all pairs X,y~X,\widetilde{\boldsymbol{y}} which satisfy (12), the choice (10) is preferable in computations, because the design matrix in (10) is sparse whenever SS is sparse, while Q1/2Q^{1/2} is dense in general.

Second, the estimator can be computed through the dual problem (15). However, the matrix Q−1Q^{-1} in (14) is dense in general, even if QQ is sparse. To avoid computing Q−1Q^{-1}, we keep β\boldsymbol{\beta} as a variable in the dual problem.

Proposition 6. Assume (6) holds, and let (β∗,u∗)(\beta^*,u^*) be a solution of

min⁡β∈Rn,u∈Rm βTQβs.t.Qβ+12DTu=Wy,0≤u≤λNI1.(17)\min_{\beta\in\mathbb{R}^n,u\in\mathbb{R}^m}\ \beta^\mathsf{T}Q\beta\quad\text{s.t.}\quad Q\beta+\frac{1}{2}D^\mathsf{T}u=Wy,\quad0\leq u\leq\lambda_{NI}\mathbf{1}. \tag*{(17)}

Then β∗=β^\beta^*=\hat{\beta}.

Proof. The constraint Qβ+12DTu=WyQ\beta+\frac{1}{2}D^\mathsf{T}u=Wy gives β=y~d−D~Tu\beta=\widetilde{y}_d-\widetilde{D}^\mathsf{T}u, and then βTQβ=∥y~d−D~Tu∥Q2\beta^\mathsf{T}Q\beta=\lVert\widetilde{y}_d-\widetilde{D}^\mathsf{T}u\rVert_Q^2. Therefore, u∗u^* is a solution to (15) and β∗=y~d−D~Tu∗\beta^*=\widetilde{y}_d-\widetilde{D}^\mathsf{T}u^*, which is equal to β^\hat{\beta} by Proposition 3.

The problem (18) contains only the matrices QQ and DD. Every row of the incidence matrix DD has two nonzero elements, and QQ is sparse whenever STSS^\mathsf{T}S is sparse. In particular, for the penalty (7) the matrix Q=W+λSDTDQ=W+\lambda_SD^\mathsf{T}D has at most n+2mn+2m nonzero elements, so it is sparse when the number of edges mm is of the order of nn, as for the chain and for the grid. In this case (18) can be solved by quadratic programming solvers for sparse problems, for example by OSQP [15], as it was done for the nearly-isotonic case in [12].

Lastly, if the constrained estimator (8) is of interest, then, by § 2, it can be obtained from (5) with a sufficiently large value of λNI\lambda_{NI}.

Conclusion and discussion

In this paper we studied the nearly-isotonic regression with respect to a general partial order, with a weighted sum of squares, where some of the weights can be equal to zero, and with an additional quadratic penalty. In our opinion, the main result of the paper is Theorem 2, where we proved that this problem can be written as a generalized lasso problem with a design matrix of full column rank and the same penalty matrix. Therefore, the estimator can be computed by the algorithms developed for the generalized lasso. Further, we derived the dual problem and wrote it as a sparse quadratic program, which can be solved by standard solvers for sparse problems. Also, we showed that for a sufficiently large penalty parameter the estimator coincides with the regularized isotonic regression.

One of the possible directions is to study the degrees of freedom of the estimator. In Theorem 2 of [21], for a design matrix of full column rank and Gaussian data with a covariance matrix proportional to the identity matrix, an unbiased estimate of the degrees of freedom of the fitted values of the generalized lasso is derived. In our Theorem 2 the design matrix has full column rank, but the fitted values Xβ^X\hat{\beta} are not the estimate of interest, and the data vector y~\widetilde{y} depends on λNI\lambda_{NI}. Moreover, even for Gaussian observations with a covariance matrix proportional to the identity matrix, the covariance matrix of y~\widetilde{y} is not of this form. Therefore, the result of [21] does not apply directly, and it is interesting to derive the degrees of freedom of the estimator β^\hat{\beta} itself, in the spirit of Theorem 6 in [12].

Another direction is to develop a path algorithm with respect to λNI\lambda_{NI}. Since the data vector y~\widetilde{y} in Theorem 2 depends on λNI\lambda_{NI}, the path algorithm of [21] does not give it directly. In particular, it is interesting to find the smallest value of λ0\lambda_0 in § 2, for which the estimator coincides with the regularized isotonic regression.

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